= Solution
The function $x^{-3}$ is strictly decreasing for positive $x$. On $a<x<a+1$, this gives $x^{-3}<a^{-3}<(x-1)^{-3}$. Integrate across that unit interval to obtain
$$
\int_a^{a+1}x^{-3}dx<a^{-3}<\int_a^{a+1}(x-1)^{-3}dx.
$$
Applying this to every integer $n\geq a$ and summing proves the <integral test> tail bounds
$$
\boxed{\frac1{2a^2}<\sum_{n=a}^\infty\frac1{n^3}<\frac1{2(a-1)^2}.}
$$
Both remain strict: the sum of the positive differences between rectangles and integrals is already bounded below by its positive first difference. Convergence of the integrals and the <p-series> makes the passage to infinity valid.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2004/paper-2-integral-comparison.png]
{title=Unit rectangles bounding a reciprocal-cube tail}
{description=Original comparison for the illustrative choice $a=3$. On the first unit interval the constant rectangle lies strictly above $x^{-3}$ and below $(x-1)^{-3}$ except at the appropriate endpoints.}
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